Solve [(x+2)^3-(x-2)^3-4(3x^2-1)]*1/10

$$[ ( x + 2 ) ^ { 3 } - ( x - 2 ) ^ { 3 } - 4 ( 3 x ^ { 2 } - 1 ) ] \cdot \frac { 1 } { 10 }$$

Evaluate

$$2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(x+2\right)^{3}.

$$\left(x^{3}+6x^{2}+12x+8-\left(x-2\right)^{3}-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-2\right)^{3}.

$$\left(x^{3}+6x^{2}+12x+8-\left(x^{3}-6x^{2}+12x-8\right)-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

To find the opposite of x^{3}-6x^{2}+12x-8, find the opposite of each term.

$$\left(x^{3}+6x^{2}+12x+8-x^{3}+6x^{2}-12x+8-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

Combine x^{3} and -x^{3} to get 0.

$$\left(6x^{2}+12x+8+6x^{2}-12x+8-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

Combine 6x^{2} and 6x^{2} to get 12x^{2}.

$$\left(12x^{2}+12x+8-12x+8-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

Combine 12x and -12x to get 0.

$$\left(12x^{2}+8+8-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

Add 8 and 8 to get 16.

$$\left(12x^{2}+16-4\left(3x^{2}-1\right)\right)\times \frac{1}{10}$$

Use the distributive property to multiply -4 by 3x^{2}-1.

$$\left(12x^{2}+16-12x^{2}+4\right)\times \frac{1}{10}$$

Combine 12x^{2} and -12x^{2} to get 0.

$$\left(16+4\right)\times \frac{1}{10}$$

Add 16 and 4 to get 20.

$$20\times \frac{1}{10}$$

Multiply 20 and \frac{1}{10} to get 2.

$$2$$

Factor

$$2$$